Question

Can someone help me set up the equation(s) for this word problem please?

A rectangular bin with an open top and volume of 38.72 cubic feet is to be built. The length of its base must be twice the width and the bin must be at least 3 feet high. Material for the base of the bin costs $12 per square foot and material for the sides costs $8 per square foot. If it costs $538.56 to build the bin, what are its dimensions?

Answer 1

yikes so many words

Answer 2

That's exactly my problem.

Answer 3

lets call the width \(x\) and that makes the length \(2x\)
the height is some unknown number \(y\) but it must be at least \(3\)
the volume is length times width times height, which will now be
\[2x^2y\]

Answer 4

so one equation we can come up with is
\[2x^2y=38.72\]

Answer 5

the base has area \(2x^2\) at a cost of \(\$12\) per square foot, so it cost \(24x^2\)

Answer 6

jeez this is a pain, but we are almost there

Answer 7

there are four sides
two have area \(xy\) and two have area \(2xy\)
a picture would help, but i can't really draw one

Answer 8

wow you're a lot faster drawer than I am. XD

Answer 9

so we get a cost of the sides of \[8xy+8xy+8\times 2xy+8\times 2xy\]
\[=48xy\]

Answer 10

for a total cost of
\[24x^2+48xy\] which we know is \(538.56\) so we can set two equations
\[24x^2+48xy=538.56\] and
\[2x^2y=38.72\]

Answer 11

i sincerely hope
a) i didn't screw this up and
b) it is making some sense to you
before we go any further, any ideas?

Answer 12

ooh i bet it is right, because this system has a nice solution !!

Answer 13

So in summary the dimensions are
2x^2(y) = 38.72 where y >/= 3
and the price is
24x^2+ 48xy = 538.56?

Answer 14

right

Answer 15

now to solve ...

Answer 16

you good with that or you need help solving it ?

Answer 17

Could you tell me at least where to start with the dimensions? I don't really know what to do first.

Answer 18

\[2x^2y=38.72\\
x^2y=19.36\\
y=\frac{19.36}{x^2}\]

Answer 19

that solves the first equation for \(y\)
now you can substitute in to the second equation

Answer 20

24x^2+ 48xy = 538.56
24x^2+ 48x(19.36/x^2) = 538.56

Answer 21

right?

Answer 22

and using that, you solve for x?

Answer 23

right

Answer 24

\[24x^2+\frac{929.28}{x}=538.56\] is what i get

Answer 25

although to be honest i just cheated at this point and asked for the answer to the system

Answer 26

24x^2 + 48x(19.36/x^2) = 538.56
24x^2 + 929.28x / 48x^3 = 538.56

Get rid of an x
x (24x + 929.28 / 48x^2) = 538.56

24x + 929.28 / 48x^2 = 538.56

Answer 27

I'm confused O.o
How did I get my x's so messed up?

Answer 28

24x^2 + 929.28x / 48x^2 = 538.56 instead of
24x^2 + 929.28x / 48x^3 = 538.56 think

Answer 29

then cancel an \(x\)

Answer 30

Oh you only cancel an x from the fraction instead from the whole side of the equation.

Answer 31

yeah
i got to \[24x^2+\frac{929.28}{x}=538.56\]

Answer 32

now i am not sure because we have to solve a cubic equation
\[24x^2-538.56x+929.28=0\]

Answer 33

no a cubic equation
\[24x^3-538.56x+929.28=0\]

Answer 34

not sure what method you are supposed to use
like i said i just cheated

Answer 35

on solution is apparently \(x=2.2\) which is the one you want because it makes \(y=4\)

Answer 36

http://www.wolframalpha.com/input/?i=24x^3-538.56x%2B929.28%3D0

Answer 37

but i just cut to the chase at the beginning
http://www.wolframalpha.com/input/?i=24x^2%2B48xy%3D538.56%2C2x^2y%3D38.72

Answer 38

Glad to see I'm not the only one using wolfram.

Answer 39

yeah i can't solve a cubic

Answer 40

So of the solutions, how do I know which is correct?

Answer 41

Do I have to plug each into the equation until I find the right one?

Answer 42

the one that has \(y=4\) since you were told that \(y>3\)

Answer 43

look at the second link

Answer 44

Oh yeah. Duh. Forgot about that.

Answer 45

Thank you so much for your help. With all those words and equations, I would have never been able to get the answer on my own. :)

Answer 46

yw
glad it worked out

Answer 47

So in the end, the dimensions would be
w= 2.2
l= 4.4
h= 4